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The Stacks project

Lemma 37.29.2. Let f : X \to Y, s : Y \to X be as in Situation 37.29.1. If g : Y' \to Y is any morphism, consider the base change diagram

\xymatrix{ X' \ar[r]_{g'} \ar[d]^{f'} & X \ar[d]_ f \\ Y' \ar@/^1pc/[u]^{s'} \ar[r]^ g & Y \ar@/_1pc/[u]_ s }

so that we obtain (X')^0 \subset X'. Then (X')^0 = (g')^{-1}(X^0).

Proof. Let y' \in Y' with image y \in Y. We may think of X^0_ y as a closed subscheme of X_ y, see for example Morphisms, Definition 29.26.3. As s(y) \in X^0_ y we conclude from Varieties, Lemma 33.7.14 that X_ y^0 is a geometrically connected scheme over \kappa (y). Hence X_ y^0 \times _ y y' \to X'_{y'} is a connected closed subscheme which contains s'(y'). Thus X_ y^0 \times _ y y' \subset (X'_{y'})^0. The other inclusion X_ y^0 \times _ y y' \supset (X'_{y'})^0 is clear as the image of (X'_{y'})^0 in X_ y is a connected subset of X_ y which contains s(y). \square


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